The answer I get is different
What is the answer you got?
I verified it now by hand, Mathematica is correct. I get the answer in terms of
$\arctan$, little simpler than what Mathematia gives
$$
I=\int\frac{\sqrt{x^{3}-1}}{x}dx
$$
Let
$u=x^{3}+1$, then
$du=3x^{2}dx$ and the above becomes
$$
I=\int\frac{\sqrt{u}}{3x^{3}}du=\frac{1}{3}\int\frac{\sqrt{u}}{u-1}dx
$$
Now let
$u=\tan^{2}v$ or
$\sqrt{u}=\tan v$, hence
$\frac{1}{2}\frac{1}
{\sqrt{u}}du=\sec^{2}vdv$ and the above becomes
\begin{align*}
I & =\frac{1}{3}\int\frac{\sqrt{u}}{\tan^{2}v-1}\left( 2\sqrt{u}\sec
^{2}v\right) dv\
& =\frac{2}{3}\int\frac{u}{\tan^{2}v-1}\sec^{2}vdv\
& =\frac{2}{3}\int\frac{\tan^{2}v}{\tan^{2}v-1}\sec^{2}vdv
\end{align*}
But
$\tan^{2}v-1=\sec^{2}v$ hence
\begin{align*}
I & =\frac{2}{3}\int\tan^{2}vdv\
& =\frac{2}{3}\left( \tan v-v\right)
\end{align*}
Substituting back
$$
I=\frac{2}{3}\left( \sqrt{u}-\arctan\left( \sqrt{u}\right) \right)
$$
Substituting back
$$
I=\frac{2}{3}\left( \sqrt{x^{3}+1}-\arctan\left( \sqrt{x^{3}+1}\right)\right)
$$